I = pressure amplitude given = 0.2 W/m²
dB = decibel reading
decibel reading from the pressure amplitude is given as
dB = 10 log₁₀ (I/10⁻¹²)
inserting the values in the above equation
dB = 10 log₁₀ (0.2/10⁻¹²)
dB = 10 log₁₀ (2 x 10⁻¹/10⁻¹²)
dB = 10 log₁₀ (2 x 10⁻¹.10¹²)
dB = 10 log₁₀ (2 x 10¹²⁻¹)
dB = 10 log₁₀ (2 x 10¹¹)
dB = 113.01 db
hence the decibel reading comes out to be 113.01 db
Solving for the acceleration of the bullet
acceleration = (vf^2 – vi^2) / 2d
acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)
acceleration = (78400 - 176400) / 0.24 m
acceleration = -98000 / 0.24
acceleration = -408333 m/s^2
Solving for contact time with board
t^2 = 2d/a
t^2 = 2 * 0.12 m / 408333 m/s^2
t^2 = 0.24 m / 408333 m/s^2
t^2 = 5.8775558 x 10^-7
t = 0.0007666 s or 767 microseconds
(I was only able to do A and B)
Please lord thank you please thank you
<span>Thallium. 81 Protons, post-transitional metal that is not found in nature. </span>
